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Leetcode025-Reverse Nodes in k-Group

Posted on Edited on In Views: Valine:
Solution Report of LeetCode Acceptted

Description

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

Follow up:

  • Could you solve the problem in O(1) extra memory space?
  • You may not alter the values in the list’s nodes, only nodes itself may be changed.

Example

Example 1:

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Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

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Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Example 3:
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Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]

Example 4:
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Input: head = [1], k = 1
Output: [1]

Constraints:

  • The number of nodes in the list is in the range sz.
  • 1 <= sz <= 5000
  • 0 <= Node.val <= 1000
  • 1 <= k <= sz

Solution

Solution 1: Recursion

  • Time Complexity:
  • Space Complexity: 
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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode curr = head;
        int count = 0;
        while (curr != null && count != k) { // find the k+1 node
            curr = curr.next;
            count++;
        }
        if (count == k) { // if k+1 node is found
            curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
            // head - head-pointer to direct part, 
            // curr - head-pointer to reversed part;
            while (count-- > 0) { // reverse current k-group: 
                ListNode tmp = head.next; // tmp - next head in direct part
                head.next = curr; // preappending "direct" head to the reversed list 
                curr = head; // move head of reversed part to a new node
                head = tmp; // move "direct" head to the next node in direct part
            }
            head = curr;
        }
        return head;
    }
}

Solution 2: Follow up, no-recursion

  • Time Complexity:
  • Space Complexity: 
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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution{
    public ListNode reverseKGroup(ListNode head, int k) {
        int n = 0;
        for (ListNode i = head; i != null; n++, i = i.next);
        
        ListNode dmy = new ListNode(0);
        dmy.next = head;
        for(ListNode prev = dmy, tail = head; n >= k; n -= k) {
            for (int i = 1; i < k; i++) {
                ListNode next = tail.next.next;
                tail.next.next = prev.next;
                prev.next = tail.next;
                tail.next = next;
            }
            
            prev = tail;
            tail = tail.next;
        }
        return dmy.next;
    }
}